Attempt:
Let $\mathcal{T}$ be the topologie generated by $\mathscr{A}$. If $\{ \mathcal{T}_{\alpha} \}$ is a collection of all topologies on $X$ that contain $\mathscr{A}$, we prove that
$$ \mathcal{T} = \bigcap_{\alpha; \mathscr{A} \subset \mathcal{T}_{\alpha}} \mathcal{T}_{\alpha} $$
First of all, we realize that intersection of topologies on X is still a topology on $X$ and it is the smallest such topology thus we always have $\bigcap \mathcal{T}_{\alpha} \subset \mathcal{T} $.
Next, Take any $U \in \mathcal{T}$. IF we can show that $U$ is in every $\mathcal{T}_{\alpha}$ then we are done.
Now, each $\mathcal{T}_{\alpha}$ is topology of $X$ which contains $\mathscr{A}$ but we dont know how the open sets look here so we may not conclude that $U$ is open with respect to these topology… BUT since $\mathcal{T}$ is generated by basis $\mathscr{A}$, then we can find some $A \in \mathscr{A} $ such that $U \subset A $ So $U \in \mathscr{A} \subset \mathcal{T}_{\alpha} $ for all $\alpha$ and ${\bf we \; are \; done}$. QED
If $\mathscr{A}$ is a subbasis, then the topology $\mathcal{T}$ generated by it is is instead the collection of all ${\bf unions}$ of finite intersections of elements of $\mathscr{A}$
. Just above we have $\bigcap \mathcal{T}_{\alpha} \subset \mathcal{T}$
so we may just show the reverse inclusion. In topology $\mathcal{T}$, the open sets are finite intersection of elements of $\mathscr{A}$ so they all lie in $\mathscr{A}$ and thus in all topologies $\mathscr{T}_{\alpha}$ and so we are basically done.
IS this correct?
Best Answer
If $\mathcal{A}$ is a base then often the topology generated by that base is the set
$$\mathcal{T}_{\mathcal{A}}:=\{\bigcup \mathcal{A}': \mathcal{A}' \subseteq \mathcal{A}\}$$
the set of all unions of subcollections of $\mathcal{A}$.
So if this is your definition, the proof for
$$\mathcal{T}_{\mathcal{A}} = \bigcap\{\mathcal{T}: \mathcal{A} \subseteq \mathcal{T}, \mathcal{T} \text{ a topology on } X\}$$
goes slightly differently: let $O$ be in $\mathcal{T}_A$ so that $O=\bigcup \mathcal{A'}$ for some $\mathcal{A'} \subseteq \mathcal{A}$. Let $\mathcal{T}$ be a topology on $X$ that contains $\mathcal{A}$. Then $\mathcal{T}$ also contains $\mathcal{A}'$ and so $O \in \mathcal{T}$ as topologies are closed under unions. As $\mathcal{T}$ was arbitrary, $O$ is in the right hand intersection, and one inclusion has been shown.
Now, $\mathcal{T}_A$ is itself a topology (because we know $\mathcal{A}$ is a base) on $X$ that contains $\mathcal{A}$, as $A = \bigcup \{A\}$ for all $A \in \mathcal{A}$, and so is one of the topologies we're intersecting on the right so that the right to left inclusion is trivial.
For a subbase $\mathcal{A}$ you can make some minor adaptations to this proof, or note that a topology contains $\mathcal{A}$ iff it contains all finite intersections of members of $\mathcal{A}$ (as all topologies are closed under finite intersections) and if we define those collections as $\mathcal{B}$ then $\mathcal{T}_{\mathcal{A}} = \mathcal{T}_{\mathcal{B}}$ which equals the intersection of all topologies that contain $\mathcal{B}$ by the previous case (where we had a base) and so also the intersection of the (same, as said !) topologies that contain $\mathcal{A}$ and we're done by a reduction to the previous case.